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Tom [10]
3 years ago
6

Consider the cell described below at 287 K: Pb | Pb2+ (1.07 M) || Fe3+ (2.13 M) | Fe Given the standard reduction potentials fou

nd on the sheet attached to the exam, calculate the cell potential after the reaction has operated long enough for the [Fe3+] to have changed by 1.052 M.
Chemistry
1 answer:
bezimeni [28]3 years ago
7 0

Answer : The cell potential is, 0.090 V

Explanation :

The values of standard reduction electrode potential of the cell are:

E^0_{[Pb^{2+}/Pb]}=-0.13V

E^0_{[Fe^{3+}/Fe]}=-0.036V

From the given cell representation we conclude that, the lead (Pb) undergoes oxidation by loss of electrons and thus act as anode. Iron (Fe) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : Pb\rightarrow Pb^{2+}+2e^-     E^0_{[Pb^{2+}/Pb]}=-0.13V

Reaction at cathode (reduction) : Fe^{3+}+3e^-\rightarrow Fe     E^0_{[Fe^{3+}/Fe]}=-0.036V

The balanced cell reaction will be,  

3Pb(s)+2Fe^{3+}(aq)\rightarrow 3Pb^{2+}(aq)+2Fe(s)

First we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

E^o=E^o_{[Fe^{3+}/Fe]}-E^o_{[Pb^{2+}/Pb]}

E^o=(-0.036V)-(-0.13V)=0.094V

Now we have to calculate the cell potential when the concentration of Fe^{3+} changed by 1.052 M.

Initial concentration of Fe^{3+} = 2.13 M

New concentration of Fe^{3+} = 2.13 - 2x

As we are given that, 2x = 1.052

So, x = 0.526

New concentration of Fe^{3+} = 2.13 - 2x = 2.13 - 1.052 = 1.078 M

Initial concentration of Pb^{2+} = 1.07 M

New concentration of Pb^{2+} = 1.07 + 3x = 1.07 + 3(0.526) = 2.648 M

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Fe^{3+}]}

n = number of electrons in oxidation-reduction reaction = 6

E_{cell} = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.094-\frac{0.0592}{6}\log \frac{(2.648)}{(1.078)}

E_{cell}=0.090V

Therefore, the cell potential is, 0.090 V

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